从列表中删除相邻的重复元素 [英] Remove adjacent duplicate elements from a list

问题描述

Google Python 类 |列表练习 -

Google Python Class | List Exercise -

给定一个数字列表，返回一个列表，其中所有相邻的 == 元素都被缩减为一个元素，所以 [1, 2, 2, 3] 返回 [1, 2, 3].您可以创建一个新列表或修改传入的列表.

Given a list of numbers, return a list where all adjacent == elements have been reduced to a single element, so [1, 2, 2, 3] returns [1, 2, 3]. You may create a new list or modify the passed in list.

我使用新列表的解决方案是 -

My solution using a new list is -

def remove_adjacent(nums):
  a = []
  for item in nums:
    if len(a):
      if a[-1] != item:
        a.append(item)
    else: a.append(item)        
  return a

这个问题甚至表明可以通过修改传入的列表来完成.但是，python 文档警告不要在使用 for 循环迭代列表时修改元素.

The question even suggests that it could be done by modifying the passed in list. However, the python documentation warned against modifying elements while iterating a list using the for loop.

我想知道除了迭代列表之外我还能尝试什么来完成这项工作.我不是在寻找解决方案，但也许是一个提示，可以将我带入正确的方向.

I am wondering what else can I try apart from iterating over the list, to get this done. I am not looking for the solution, but maybe a hint that can take me into a right direction.

更新

-使用建议的改进更新了上述代码.

-updated the above code with suggested improvements.

- 使用建议的提示在 while 循环中尝试以下操作 -

-tried the following with a while loop using suggested hints -

def remove_adjacent(nums):
  i = 1
  while i < len(nums):    
    if nums[i] == nums[i-1]:
      nums.pop(i)
      i -= 1  
    i += 1
  return nums

推荐答案

使用生成器迭代列表中的元素，并yield只有在它发生变化时才产生一个新的.

Use a generator to iterate over the elements of the list, and yield a new one only when it has changed.

itertools.groupby 确实如此这个.

itertools.groupby does exactly this.

如果你迭代一个副本，你可以修改传入的列表:

You can modify the passed-in list if you iterate over a copy:

for elt in theList[ : ]:
    ...

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